Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = \dfrac{1}{3}x + 1}\enspace$ and passes through the point ${(3, -6)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${\dfrac{1}{3}}$ , and its negative reciprocal is ${-3}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -3x + b}\enspace$ We can plug our point, $(3, -6)$ , into this equation to solve for ${b}$ , the y-intercept. $-6 = {-3}(3) + {b}$ $-6 = -9 + {b}$ $-6 + 9 = {b} = 3$ The equation of the perpendicular line is $\enspace {y = -3x + 3}\enspace$. ${m = -3, \enspace b = 3}$